#+PROPERTY: header-args :session gaussian_analytic :kernel python :pandoc yes :async yes

In this notebook, we compare the analytic solution with the numeric
result.  A nicer version with the code outsourced into a proper python
module can be found [[file:analytic_calculation_with_gaussflow.org][here]].

#+begin_src jupyter-python :results none
  import scipy
  import numpy as np
  import matplotlib.pyplot as plt
  from scipy import integrate
  import sys
  import utilities
  import quadpy
  from hops.util import bcf
  import itertools
#+end_src

* System Set-Up
#+begin_src jupyter-python :results none
  Ω = 1.5
  A = np.array([[0, Ω], [-Ω, 0]])
  η = 2
  ω_c = 1
  t_max = 25

  def α(t):
      return η / np.pi * (ω_c / (1 + 1j * ω_c * t)) ** 2


  def α_0_dot(t):
      return 2 * η / (1j * np.pi) * (ω_c / (1 + 1j * ω_c * t)) ** 3
#+end_src

* Exponential Fit
First we need an exponential fit for our BCF.
#+begin_src jupyter-python
  W, G_raw = utilities.fit_α(α, 3, 80, 10_000)
  L, P = utilities.fit_α(α_0_dot, 5, 80, 10_000)
#+end_src

#+RESULTS:

Looks quite good.
#+begin_src jupyter-python
  τ = np.linspace(0, t_max, 1000)
  with utilities.hiro_style():
      fig, ax = utilities.plot_complex(τ, α(τ), label="exact")
      utilities.plot_complex(
          τ, utilities.α_apprx(τ, G_raw, W), ax=ax, label="fit", linestyle="--"
      )
#+end_src

#+RESULTS:
[[file:./.ob-jupyter/9c454eedaf97fa770850f6d4948856df8948b8df.svg]]


#+begin_src jupyter-python
  τ = np.linspace(0, -t_max, 1000)
  with utilities.hiro_style():
      fig, ax = utilities.plot_complex(τ, α(τ), label="exact")
      utilities.plot_complex(
          τ, utilities.α_apprx(-τ, G_raw.conj(), W.conj()), ax=ax, label="fit", linestyle="--"
      )
#+end_src

#+RESULTS:
[[file:./.ob-jupyter/37cc788143d55668dd7087df2e59b1e330223895.svg]]

#+begin_src jupyter-python
  fig, ax = utilities.plot_complex(τ, α_0_dot(τ), label="exact")
  utilities.plot_complex(
      τ, utilities.α_apprx(τ, P, L), ax=ax, label="fit", linestyle="--"
  )
#+end_src

#+RESULTS:
:RESULTS:
| hline | <AxesSubplot:> |
[[file:./.ob-jupyter/666257d29aba72e31b8e20527e5cf43a8c475808.svg]]
:END:

* Analytical Solution
** Calculate the Magic Numbers
We begin with the $\varphi_n$ and $G_n$ from the original ~G~.
#+begin_src jupyter-python
  φ = -np.angle(G_raw)
  φ
#+end_src

#+RESULTS:
: array([-1.18710245,  0.64368323,  2.11154332])

#+begin_src jupyter-python
  G = np.abs(G_raw)
  G
#+end_src

#+RESULTS:
: array([0.51238635, 0.62180167, 0.10107935])

Now we calculate the real and imag parts of the ~W~ parameters and
call them $\gamma_n$ and $\delta_n$.

#+begin_src jupyter-python
  γ, δ = W.real, W.imag
#+end_src

#+RESULTS:

Now the \(s_n, c_n\).
#+begin_src jupyter-python
  s, c = np.sin(φ), np.cos(φ)
#+end_src

#+RESULTS:

Now we calculate the roots of $f_n(z)=-G_n ((z+\gamma_n) s_n + \delta_n
c_n)$.
Normally we should be vary of one of the \(\deltas\) being zero, but
this is not the case.
#+begin_src jupyter-python
  roots_f = -(γ + δ * c/s)
  roots_f
#+end_src

#+RESULTS:
: array([-1.19725058, -2.0384181 , -0.23027627])

Now the \(z_k\) the roots of \(\delta_k^2 + (\gamma_k + z)^2\). *We
don't include the conjugates.*
#+begin_src jupyter-python
  roots_z = -W
  roots_z
#+end_src

#+RESULTS:
: array([-2.29230292-2.71252483j, -1.07996581-0.719112j  ,
:        -0.28445344-0.09022829j])

** Construct the Polynomials
#+begin_src jupyter-python
  from numpy.polynomial import Polynomial
#+end_src

#+RESULTS:

We later need \(f_0(z) = \prod_k (z-z_k) (z-z_k^{\ast})\).
#+begin_src jupyter-python
  f_0 = utilities.poly_real(Polynomial.fromroots(np.concatenate((roots_z, roots_z.conj()))))
  f_0
#+end_src

#+RESULTS:

Another polynomial is simply \(p_1 = (z^2 + \Omega^2)\prod_k (z-z_k)
(z-z_k^{\ast})\) and we can construct it from its roots.
#+begin_src jupyter-python
  p_1 = Polynomial([Ω**2, 0, 1]) * f_0
  p_1
#+end_src

#+RESULTS:

The next ones are given through \(q_n=\Omega f_n(z) \prod_{k\neq
n}(z-z_k) (z-z_k^{\ast})\)
#+begin_src jupyter-python
  q = [
      -G_c
      ,* Ω * s_c
      ,* utilities.poly_real(Polynomial.fromroots(
          np.concatenate(
              (
                  [root_f],
                  utilities.except_element(roots_z, i),
                  utilities.except_element(roots_z, i).conj(),
              )
          )
      ))
      for root_f, G_c, γ_c, δ_c, s_c, c_c, i in zip(roots_f, G, γ, δ, s, c, range(len(c)))
  ]
#+end_src

#+RESULTS:

With this we construct our master polynomial \(p = p_1 + \sum_n q_n\).
#+begin_src jupyter-python
  p = p_1 + sum(q)
  p
#+end_src

#+RESULTS:

And find its roots.
#+begin_src jupyter-python
  master_roots = p.roots()
  master_roots
#+end_src

#+RESULTS:
: array([-2.28139877-2.68284887j, -2.28139877+2.68284887j,
:        -0.93979297-0.62025112j, -0.93979297+0.62025112j,
:        -0.24204514-0.10390896j, -0.24204514+0.10390896j,
:        -0.19348529-1.49063362j, -0.19348529+1.49063362j])

Let's see if they're all unique. This should make things easier.
#+begin_src jupyter-python
  np.unique(master_roots).size == master_roots.size
#+end_src

#+RESULTS:
: True

Very nice!

** Calculate the Residuals
These are the prefactors for the diagonal.
#+begin_src jupyter-python
  R_l = f_0(master_roots) / p.deriv()(master_roots)
  R_l
#+end_src

#+RESULTS:
: array([ 0.00251589-0.00080785j,  0.00251589+0.00080785j,
:         0.06323421+0.02800137j,  0.06323421-0.02800137j,
:         0.02732669-0.00211665j,  0.02732669+0.00211665j,
:        -0.09307679+0.36145133j, -0.09307679-0.36145133j])

And the laplace transform of \(\alpha\).
#+begin_src jupyter-python
  def α_tilde(z):
      return (
          -G[None, :]
          ,* ((z[:, None] + γ[None, :]) * s[None, :] + δ[None, :] * c[None, :])
          / (δ[None, :] ** 2 + (γ[None, :] + z[:, None]) ** 2)
      ).sum(axis=1)
#+end_src

#+RESULTS:

And these are for the most compliciated element.
#+begin_src jupyter-python
  R_l_21 = (Ω + α_tilde(master_roots))* f_0(master_roots) / p.deriv()(master_roots)
  R_l_21
#+end_src

#+RESULTS:
: array([-0.00325014-0.02160514j, -0.00325014+0.02160514j,
:         0.00074814-0.05845205j,  0.00074814+0.05845205j,
:        -0.00094159-0.00084894j, -0.00094159+0.00084894j,
:         0.00344359+0.56219924j,  0.00344359-0.56219924j])


Now we can calculate \(G\).
#+begin_src jupyter-python
  def G_12_ex(t):
      t = np.asarray(t)
      t_shape = t.shape

      if len(t.shape) == 0:
          t = t.reshape((1,))
      return Ω * (R_l[None, :] * np.exp(t[:, None] * master_roots[None, :])).real.sum(
          axis=1
      ).reshape(t_shape)


  def G_11_ex(t):
      t = np.asarray(t)
      t_shape = t.shape
      if len(t.shape) == 0:
          t = np.array([t])

      return (
          R_l[None, :]
          ,* master_roots[None, :]
          ,* np.exp(t[:, None] * master_roots[None, :])
      ).real.sum(axis=1).reshape(t_shape)


  def G_21_ex(t):
      t = np.asarray(t)
      return -(R_l_21[None, :] * np.exp(t[:, None] * master_roots[None, :])).real.sum(
          axis=1
      )


  def G_21_ex_alt(t):
      t = np.asarray(t)
      return (
          R_l[None, :]
          ,* master_roots[None, :] ** 2
          ,* np.exp(t[:, None] * master_roots[None, :])
      ).real.sum(axis=1) / Ω


  def G_ex(t):
      t = np.asarray(t)
      if t.size == 1:
          t = np.array([t])
      diag = G_11_ex(t)
      return (
          np.array([[diag, G_12_ex(t)], [G_21_ex(t), diag]]).swapaxes(0, 2).swapaxes(1, 2)
      )


  def G_ex_new(t):
      t = np.asarray(t)
      if t.size == 1:
          t = np.array([t])

      g_12 = R_l[None, :] * np.exp(t[:, None] * master_roots[None, :])
      diag = master_roots[None, :] * g_12
      g_21 = master_roots[None, :] * diag / Ω

      return (
          np.array([[diag, g_12 * Ω], [g_21, diag]])
          .real.sum(axis=3)
          .swapaxes(0, 2)
          .swapaxes(1, 2)
      )
#+end_src

#+RESULTS:

#+begin_src jupyter-python
  with utilities.hiro_style():
      plt.plot(τ, G_ex_new(τ).reshape(len(τ), 4))
#+end_src

#+RESULTS:
[[file:./.ob-jupyter/43540f787051db4b09fa82a1204956050edca117.svg]]

* Energy Flow
We adopt the terminology from my notes.
#+begin_src jupyter-python :results none
  A = G_11_ex
  B = G_12_ex
  Pk = P
  Bk = R_l * Ω
  Gk = G_raw
  Lk = L
  Ck = -master_roots # Mind the extra sign!
  Wk = W
  Ak = master_roots * R_l
  n = 1
  q_s_0 = 1 + 2 * n
  p_s_0 = 1 + 2 * n
  qp_0 = 1j
#+end_src

First, we calculate the coefficient matrices.
#+begin_src jupyter-python :results none
  Γ1 = (Bk[:, None] * Pk[None, :]) / (Lk[None, :] - Ck[:, None])
  Γ2 = (Bk[:, None] * Gk[None, :]) / (Ck[:, None] - Wk[None, :])
  Γ3 = (Bk[:, None] * Gk.conj()[None, :]) / (Ck[:, None] + Wk.conj()[None, :])
  ΓA = (Ak[:, None] * Pk[None, :]) / (Lk[None, :] - Ck[:, None])
#+end_src

The convolution free part is easy to calculate.
#+begin_src jupyter-python :results none
  def A_conv(t):
      t = np.asarray(t)
      result = np.zeros_like(t, dtype="complex128")
      for (n, m) in itertools.product(range(len(Ak)), range(len(Pk))):
          result += ΓA[n, m] * (np.exp(-Ck[n] * t) - np.exp(-Lk[m] * t))

      return result


  def B_conv(t):
      t = np.asarray(t)
      result = np.zeros_like(t, dtype="complex128")
      for (n, m) in itertools.product(range(len(Bk)), range(len(Pk))):
          result += Γ1[n, m] * (np.exp(-Ck[n] * t) - np.exp(-Lk[m] * t))

      return result


  def flow_conv_free(t):
      a, b = A(t), B(t)
      ac, bc = A_conv(t), B_conv(t)
      return (
          -1
          / 2
          ,* (
              q_s_0 * a * ac + p_s_0 * b * bc + qp_0 * a * bc + qp_0.conjugate() * b * ac
          ).imag
      )
#+end_src


The part containing the convolution with the BCF is a bit mode
involved.  The first version originates from my calculations and the
second one arose from sagemath. Thank God! They do aggree now.
#+begin_src jupyter-python :results none
  def conv_part(t):
      t = np.asarray(t)
      result = np.zeros_like(t, dtype="float")
      for (m, k, n, l) in itertools.product(
          range(len(Bk)), range(len(Pk)), range(len(Bk)), range(len(Gk))
      ):
          g_1_2 = (
              Γ1[m, k]
              ,* Γ2[n, l]
              ,* (
                  (1 - np.exp(-(Ck[m] + Wk[l]) * t)) / (Ck[m] + Wk[l])
                  + (np.exp(-(Ck[m] + Ck[n]) * t) - 1) / (Ck[m] + Ck[n])
                  + (np.exp(-(Lk[k] + Wk[l]) * t) - 1) / (Lk[k] + Wk[l])
                  + (1 - np.exp(-(Lk[k] + Ck[n]) * t)) / (Lk[k] + Ck[n])
              )
          )

          g_1_3 = (
              Γ1[m, k]
              ,* Γ3[n, l]
              ,* (
                  (1 - np.exp(-(Ck[m] + Ck[n]) * t)) / (Ck[m] + Ck[n])
                  - (1 - np.exp(-(Lk[k] + Ck[n]) * t)) / (Lk[k] + Ck[n])
                  - (np.exp(-(Ck[n] + Wk[l].conj() * t)) - np.exp(-(Ck[m] + Ck[n]) * t))
                  / (Ck[m] - Wk[l].conj())
                  + (np.exp(-(Ck[n] + Wk[l].conj() * t)) - np.exp(-(Lk[k] + Ck[n]) * t))
                  / (Lk[k] - Wk[l].conj())
              )
          )

          result += -1 / 2 * (g_1_2.imag + g_1_3.imag)

      return result


  import monster_formula


  def conv_part_sage(t):
      t = np.asarray(t)
      result = np.zeros_like(t, dtype="float")
      for (n, k, l, m) in itertools.product(
          range(len(Bk)), range(len(Pk)), range(len(Gk)), range(len(Bk))
      ):

          result += (
              -1
              / 2
              ,* (
                  monster_formula.conv_part(
                      t,
                      Pk[k],
                      Lk[k],
                      Bk[n],
                      Ck[n],
                      Bk[m],
                      Ck[m],
                      Gk[l],
                      Wk[l],
                      Gk[l].conj(),
                      Wk[l].conj(),
                  )
              ).imag
          )

      return result
#+end_src

The contribution of the convolution part is pretty minor for zero
temperature.
#+begin_src jupyter-python
  τ = np.linspace(0, t_max, 5000)
  plt.plot(τ, conv_part(τ))
#+end_src

#+RESULTS:
:RESULTS:
| <matplotlib.lines.Line2D | at | 0x7f85099ac190> |
[[file:./.ob-jupyter/ed8ce474b528ddf7104399b40e018db0180b3ee5.svg]]
:END:

Now we can define the flow:
#+begin_src jupyter-python :results none
  def flow(t):
      return flow_conv_free(t) + conv_part(t)
#+end_src

And plot it against the numerically obtained flow.
#+begin_src jupyter-python
  with open('res.npy', 'rb') as f:
      old_τ = np.load(f)
      old_flow_τ = np.load(f)
  with utilities.hiro_style():
      plt.plot(old_τ, flow(old_τ))
      plt.plot(old_τ, old_flow_τ)
#+end_src

#+RESULTS:
[[file:./.ob-jupyter/6c77bc1a9d844fa3e33a14c077310c32a7eec4a8.svg]]

They do aggree but I think the accuracy of the numerical integration
is getting worse with time.

#+begin_src jupyter-python
  scipy.integrate.quad(flow, 0, 20)
#+end_src

#+RESULTS:
| 1.6004236143630652 | 1.564508138231918e-08 |

Integresting... maybe the fact that it does not converge to zero
exactly has to do with this...